Sequence And Series Question 238
Question: In the four numbers first three are in G.P. and last three are in A.P. whose common difference is 6. If the first and last numbers are same, then first will be
[IIT 1974]
Options:
A) 2
B) 4
C) 6
D) 8
Show Answer
Answer:
Correct Answer: D
Solution:
Let the numbers be $ \frac{a}{r},\ a,\ ar,\ 2ar-a $ ?..(i) Where first three numbers are in G.P. and last three are in A.P. Given that the common difference of A.P. is 6, so $ ar-a=6 $ ?..(ii) Also given $ \frac{a}{r}=2ar-a\Rightarrow \frac{a}{r}=2,(ar-a)+a $
$ \Rightarrow $ $ \frac{a}{r}=2(6)+a, $ from (ii)
$ \Rightarrow $ $ ( \frac{a}{r} )-a=12 $
$ \Rightarrow $ $ a(1-r)=12r $
$ \Rightarrow $ $ r=-\frac{1}{2} $ From (i) we get, $ a[ ( -\frac{1}{2} )-1 ]=6 $ or $ a=-4 $ Required numbers from (i) are $ 8,\ -4,\ 2,\ 8 $ .
BETA
